Find the higher order derivative of sin(x^3+y^5)

\frac{d^2}{dx^2}\left(\sin\left(x^3+y^5\right)\right)

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Answer

$3\left(2x\cos\left(y^5+x^3\right)-3x^{4}\sin\left(y^5+x^3\right)\right)$

Step by step solution

Problem

$\frac{d^2}{dx^2}\left(\sin\left(x^3+y^5\right)\right)$
1

Rewriting the high order derivative

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\frac{d}{dx}\left(\sin\left(y^5+x^3\right)\right)\right)$
2

The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if ${f(x) = \sin(x)}$, then ${f'(x) = \cos(x)\cdot D_x(x)}$

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\frac{d}{dx}\left(y^5+x^3\right)\cos\left(y^5+x^3\right)\right)$
3

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\left(\frac{d}{dx}\left(y^5\right)+\frac{d}{dx}\left(x^3\right)\right)\cos\left(y^5+x^3\right)\right)$
4

The derivative of the constant function is equal to zero

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\left(0+\frac{d}{dx}\left(x^3\right)\right)\cos\left(y^5+x^3\right)\right)$
5

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{d^{\left(2-1\right)}}{dx^{\left(2-1\right)}}\left(\left(0+3x^{2}\right)\cos\left(y^5+x^3\right)\right)$
6

Subtract the values $2$ and $-1$

$\frac{d^{1}}{dx^{1}}\left(\left(0+3x^{2}\right)\cos\left(y^5+x^3\right)\right)$
7

$x+0=x$, where $x$ is any expression

$\frac{d^{1}}{dx^{1}}\left(3x^{2}\cos\left(y^5+x^3\right)\right)$
8

Any expression to the power of $1$ is equal to that same expression

$\frac{d}{dx}\left(3x^{2}\cos\left(y^5+x^3\right)\right)$
9

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$3\frac{d}{dx}\left(x^{2}\cos\left(y^5+x^3\right)\right)$
10

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=x^{2}$ and $g=\cos\left(y^5+x^3\right)$

$3\left(x^{2}\cdot\frac{d}{dx}\left(\cos\left(y^5+x^3\right)\right)+\cos\left(y^5+x^3\right)\frac{d}{dx}\left(x^{2}\right)\right)$
11

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$3\left(x^{2}\cdot\frac{d}{dx}\left(\cos\left(y^5+x^3\right)\right)+2x\cos\left(y^5+x^3\right)\right)$
12

The derivative of the cosine of a function is equal to minus the sine of the function times the derivative of the function, in other words, if $f(x) = \cos(x)$, then $f'(x) = -\sin(x)\cdot D_x(x)$

$3\left(2x\cos\left(y^5+x^3\right)-x^{2}\cdot\frac{d}{dx}\left(y^5+x^3\right)\sin\left(y^5+x^3\right)\right)$
13

The derivative of a sum of two functions is the sum of the derivatives of each function

$3\left(2x\cos\left(y^5+x^3\right)-x^{2}\left(\frac{d}{dx}\left(y^5\right)+\frac{d}{dx}\left(x^3\right)\right)\sin\left(y^5+x^3\right)\right)$
14

The derivative of the constant function is equal to zero

$3\left(2x\cos\left(y^5+x^3\right)-x^{2}\left(0+\frac{d}{dx}\left(x^3\right)\right)\sin\left(y^5+x^3\right)\right)$
15

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$3\left(2x\cos\left(y^5+x^3\right)-x^{2}\left(0+3x^{2}\right)\sin\left(y^5+x^3\right)\right)$
16

$x+0=x$, where $x$ is any expression

$3\left(2x\cos\left(y^5+x^3\right)-1\cdot 3x^{2}x^{2}\sin\left(y^5+x^3\right)\right)$
17

Multiply $3$ times $-1$

$3\left(2x\cos\left(y^5+x^3\right)-3x^{2}x^{2}\sin\left(y^5+x^3\right)\right)$
18

When multiplying exponents with same base we can add the exponents

$3\left(2x\cos\left(y^5+x^3\right)-3x^{4}\sin\left(y^5+x^3\right)\right)$

Answer

$3\left(2x\cos\left(y^5+x^3\right)-3x^{4}\sin\left(y^5+x^3\right)\right)$

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.29 seconds

Views:

100